search for: medve

Dear All, I have data which contains 14 variables. And I have to regress one of variables on each variable (simple 13 linear regressions) I try to make a loop and store only R-squared colnames(boston) [1] "CRIM" "ZN" "INDUS" "CHAS" "NOX" "RM" "AGE" [8] "DIS" "RAD"

I am having problem using the step function for a linear regression model. I've created an initial model containing only the intercept. Then using the step function, I've selected three variables to be considered for the model. > x0.lm<- lm(MEDV~1, data = x) > > anova(x0.lm) Analysis of Variance Table Response: MEDV Df Sum Sq Mean Sq F value Pr(>F)

> >  Dear moderators, Please help me encode the program instructed by follows. Thank u! Apply the methods introduced in Sections 4.2.1 and 4.2.2, say the > rank-based variable selection and BIC criterions, to the Boston housing > data. >  The Boston housing data contains 506 observations, and is publicly available in the R package mlbench (dataset “BostonHousing”).  The

Dear all, I'm having some problems with using the step() function inside another function. I think it is an environment problem but I do not know how to overcome it. Any suggestions are appreciated. I've prepared a simple example to illustrate my problem: > library(MASS) > data(Boston) > my.fun <- function(dataset) { + l <- lm(medv ~ .,data=dataset) + final.l <-

On Wed, 29 Nov 2000, Matt Pocernich wrote: > I am having problem using the step function for a linear regression model. I've created an initial model containing only the intercept. Then using the step function, I've selected three variables to be considered for the model. > > > > x0.lm<- lm(MEDV~1, data = x) > > > > anova(x0.lm) > Analysis of

On Wed, 29 Nov 2000, Matt Pocernich wrote: > I am having problem using the step function for a linear regression model. I've created an initial model containing only the intercept. Then using the step function, I've selected three variables to be considered for the model. > > > > x0.lm<- lm(MEDV~1, data = x) > > > > anova(x0.lm) > Analysis of

According to the man page for formula, "a formula object has an associated environment". However, update.default doesn't use this environment, which creates problems like the following: make.model <- function(x) { lm(medv~.,x) } library(MASS) data(Boston) fit = make.model(Boston) fit = update(fit,".~.-crim") # Object "x" not found Here is a

Hello, I'm currently dealing with a rather strange problem when using the function "stepAIC" ("MASS" package). The setting is the following: From model learning data sets ("learndata"), I want to be able to build prediction functions (in order to save them in a file for further use). This is done by the function "pred.function" (see below). Therein,

This question concerns rpart's facility for user-defined functions that accomplish splitting. I was interested in modifying the code so that in each terminal node, a linear regression is fit to the data. It seems that from the allowable inputs in the user-defined functions, that this may not be possible, since they have the form: function(y, wt, parms) (in the case of the

Hi, I am having problems passing arguments to method="gbm" using the train() function. I would like to train gbm using the laplace distribution or the quantile distribution. here is the code I used and the error: gbm.test <- train(x.enet, y.matrix[,7], method="gbm", distribution=list(name="quantile",alpha=0.5), verbose=FALSE,

Dear list members, I am trying to get information on how to fit a linear regression with constrained parameters. Specifically, I have 8 predictors , their coeffiecients should all be non-negative and add up to 1. I understand it is a quadratic programming problem but I have no experience in the subject. I searched the archives but the results were inconclusive. Could someone provide suggestions

I'm looking for guidance on how to implement forward stepwise regression using lmStepAIC in Caret. The stepwise "direction" appears to default to "backward". When I try to use "scope" to provide a lower and upper model, Caret still seems to default to "backward". Any thoughts on how I can make this work? Here is what I tried: itemonly <-

Hello everyone, I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. I want to do a constrained regression such that a>0 and (1-a) >0 for the model: Y = a*X1 + (1-a)*X2 I tried the help on the constrained regression in R but I concede that it was not helpful. Any help is greatly appreciated -- Thanks, Jim. [[alternative HTML version deleted]]

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) Please: Show me the tree. Mark -------- Original Message -------- Subject: Re: [R] help with rpart From: "Stephen Milborrow" <[1]milbo at sonic.net>

Hello Tom, Tom Stellard via llvm-dev <llvm-dev <at> lists.llvm.org> writes: > > Hi, > > I'm working on uploading all the packages now. I've downloaded them files several times from different locations (suspected my webgw might muck with the content) using different clients and the result is the same: $ gpg --verify ~/Downloads/cfe-3.8.1.src.tar.xz.sig

FWIW, I wrote a little function to extract variable importance as defined in the CART book a while ago. It's rather limited: Only works for regression problem, and you need to set maxsurrogate=0 and maxcompete=0. It may (or may not) help you: varimp.rpart <- function(x) { dev <- x$frame[, c("var", "dev")] dev <- dev[dev$var != "<leaf>",

Cubist is a rule-based machine learning model for regression. Parts of the Cubist model are described in: Quinlan. Learning with continuous classes. Proceedings of the 5th Australian Joint Conference On Artificial Intelligence (1992) pp. 343-348 Quinlan. Combining instance-based and model-based learning. Proceedings of the Tenth International Conference on Machine Learning

Cubist is a rule-based machine learning model for regression. Parts of the Cubist model are described in: Quinlan. Learning with continuous classes. Proceedings of the 5th Australian Joint Conference On Artificial Intelligence (1992) pp. 343-348 Quinlan. Combining instance-based and model-based learning. Proceedings of the Tenth International Conference on Machine Learning

...edora release 11 (Leonidas) ext4 protocols: imap pop3 listen: * ssl: no login_dir: /var/run/dovecot/login login_executable(default): /usr/libexec/dovecot/imap-login login_executable(imap): /usr/libexec/dovecot/imap-login login_executable(pop3): /usr/libexec/dovecot/pop3-login login_greeting: PREVED MEDVED first_valid_uid: 8 last_valid_uid: 8 first_valid_gid: 12 last_valid_gid: 12 mail_location: maildir:/home/email/%d/%n mail_debug: yes mail_executable(default): /usr/libexec/dovecot/imap mail_executable(imap): /usr/libexec/dovecot/imap mail_executable(pop3): /usr/libexec/dovecot/pop3 mail_plugin_dir...

Hello, Is the first time I am using SNOW package and I am trying to tune the cost parameter for a linear SVM, where the cost (variable cost1) takes 10 values between 0.5 and 30. I have a large dataset and a pc which is not very powerful, so I need to tune the parameters using both CPUs of the pc. Somehow I cannot manage to do it. It seems that both CPUs are fitting the model for the same values